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f1-32=(9/5)c1
f2-32=(9/5)c2
for increase, f1-f2=(9/5)(c-c2)

#include<stdio.h>
double increase(int d)
{
    double c=(double)5*d/9;
    return c;
}
 
int main()
{
    int tc,f,c,i=1;
    double result;
    scanf("%d",&tc);
    while(tc--)
    {
        scanf("%d%d",&c,&f);
        result=(double)c+increase(f);
        printf("Case %d: %.2lf\n",i,result);
        i++;
    }
    return 0;
}

 

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